I think he's using the method we all learned in high school: to get an inverse function, replace x with y and vice-versa, then solve for y (where y, in this case, is really [itex]y^{-1}(x)[/itex] ). For instance:
[tex]y = e^{x}[/tex]
Repleace x with y
[tex]x = e^y[/tex]
Solve for y
[tex]\ln{x} = \ln{e^y} = y[/tex]
Therefore, the inverse of [itex]y=e^x[/itex] is [itex]y=\ln{x}[/itex]
I doubt he's up to the point of using Power Series expansions (most people learn that after calculus, and this is the precalculus forum) or matrices.
That said, this one is tricky and I don't recall how you'd do it aside from the more complicated methods mentioned here.

ok... so g'(x) = 1/ 1 + cos(g(x))
now they want g'(pi)...and im stuck again, i cant find any trig identities that can help me in g'(x)

Remember that if f(x) and g(x) are inverse functions then
[tex] g'(x) = \frac 1 {f'(g(x))} [/tex]
and if
f(x) = y
then
g(y) = x
so to find g'(x) at a point say t, g'(t), just solve f(x) = t for x then that value of x you find is g(t) then to find the derivative at that point just take 1 over the derivative of f evaluated at the x you found.

No, f(x)=[itex]\pi[/itex], so
[tex]\pi=x + \sin{x}[/tex]
Can you think of the value of x that makes this true? This value is [itex]g(\pi)[/itex] because, by definition of the inverse function, if
[tex]f(a)=b[/tex]
then
[tex]f^{-1}(b)=a[/tex]
So then what you want to do is find f'(x) and then sub in the value of [itex]g(\pi)[/itex] that you get to find [itex]f'(g(\pi))[/itex]. You know where to go from there.